A naive question about SGA4

Question

Can someone explain to me the meaning of remark 1.1.2 at the begining of SGA4.1?

It says that if $C$ is a category that belongs to some universe $U$ (which I understand as "$\mathrm{Ob}(C)$ and $\mathrm{Mor}(C)$ are elements of $U$, or $U$-sets"), then the functor category $\mathrm{Fonct}(C,U\mbox{-}\mathrm{Set})$ from $C$ to the category of $U$-sets (whose objects are the elements of $U$ and whose morphisms are applications between these) has neither of the following two properties:

1. $\mathrm{Fonct}(C,U\mbox{-}\mathrm{Set})$ is a subset of $U$.

2. for any two functors $F,G\in \mathrm{Fonct}(C,U\mbox{-}\mathrm{Set})$, the set $\mathrm{Hom}(F,G)$ is a member of $U$.

For this reason Grothendieck introduces a more general notion of '$U$-category' as a category whose hom sets are not necessarily members of $U$ but are all equipotent to members of $U$.

My problem is that I don't see how $\mathrm{Fonct}(C,U\mbox{-}\mathrm{Set})$ can fail to have properties 1 and 2.

If $C$ is a member of $U$, then any functor from $C$ to $U$-Set can be seen as a subset of a member of $U$ (or a pair of such subsets), so is a member of $U$, which implies that $\mathrm{Ob}(\mathrm{Fonct}(C,U\mbox{-}\mathrm{Set}))$ is a subset of $U$; and likewise, if $F,G$ are functors from $C$ to $U$-Set, a morphism from $F$ to $G$ will be a function from $C$ (a member of $U$) to the union (over $X\in C$) of the sets defined by the morphisms $F(X)\rightarrow G(X)$, which are members of $U$.

Likely I did not read this remark right or there is something wrong in the above reasoning... Many thanks in advance for any help in clarifying this.

Answer 1

Let me explain this in detail, using the usual Grothendieck/Bourbaki foundations (cf. the references given in my above comment).

Consider a universe $U$, the category $D$ of $U$-sets, a category $C$, and the category $Hom(C,D)$ of functors from $C$ to $D$.

Claim 1: $Ob(Hom(C,D))\not\subseteq U$.

Proof: Consider a functor $F\colon C\rightarrow D$. We will show that $F\notin U$. Such a functor $F$ is, by definition, a pair $(F_0,F_1)$ consisting of maps $F_0\colon Ob(C)\rightarrow Ob(D)$ and $F_1\colon Mor(C)\rightarrow Mor(D)$ fulfilling certain properties. So, $F\notin U$ if and only if $F_0\notin U$ or $F_1\notin U$. We will show that $F_0\notin U$. By definition, a map $F_0\colon Ob(C)\rightarrow Ob(D)$ is a triple $(F_0',Ob(C),Ob(D))$, where $F_0'$ is a certain subset of $Ob(C)\times Ob(D)$. So, we have $F_0\notin U$ if and only if at least one of the components of this triple is not an element of $U$. But since $Ob(D)=U$ and $U\notin U$, this is fulfilled and the claim is proven. (Note that a universe is never an element of itself.)

Consider now in addition two functors $F,G\colon C\rightarrow D$, and the set $Hom(F,G)$ of morphisms of functors from $F$ to $G$ (in the category $Hom(C,D)$).

Claim 2: $Hom(F,G)\notin U$.

Proof: If $Hom(F,G)$ is an element of $U$, then so is every element of the set $Hom(F,G)$. Thus, it suffices to show that elements of $Hom(F,G)$ are not elements of $U$. Let $a\in Hom(F,G)$. Then, by definition, $a$ is a map $Ob(C)\rightarrow Mor(D)$, hence a triple $(a',Ob(C),Mor(D))$, and it suffices to show that one of its components is not an element of $U$. We will show that $Mor(D)\notin U$. Indeed, if we assume to the contrary that $Mor(D)\in U$, then so is the union $\bigcup_{f\in Mor(D)}pr_3(f)$, where the third projection of a map is its codomain. Now, this union equals the union of all elements of $U$ and is thus equal to $U$. We arrive again at the contradiction that $U\in U$. Herewith the claim is proven.

Answer 2

In retrospect, it appears that claims 1 and 2 both lie on the fact that one normally insists, in a category C (specifically, in the category of sets), that a morphism should determine its domain and codomain. Equivalently, the hom-sets hom(A,B) and hom(A',B') of a category are assumed to be disjoint if (A,B) is distinct of (A',B').

To quote the begining of Grothendieck's Tohoku paper (where he defines a category):

"Enfin, il sera prudent de supposer que la donnee d'un morphisme u determine ses objets de "depart" et "d'arrivee", en d'autres termes que si (A,B) et (A',B') sont deux couples distincts d'objets de C, alors Hom(A,B) et Hom (A',B') sont deux ensembles disjoints."

While it seems clearly desirable to make this assumption, I wonder if this is more than a technical convenience, or if there is really something fundamental to it. It is also clear from the above quote that one could imagine doing otherwise. For example, one could define a function between two sets E,F as being just a subset of ExF having the property of being a functional graph. Then none of the claims 1 and 2 (which I cannot help but to perceive as artificial) would have to hold anymore.